To maximize the volume of the open-top box, we need to determine the dimensions after cutting and folding. Let $x$ represent the length of each square cut from each corner. After cutting out these squares, the box's base dimensions and height will be modified based on $x$.

1. Determine the box dimensions after cutting:

   • The initial dimensions of the cardboard are $9$ feet (length) and $8$ feet (width).
   • After cutting out squares of side length $x$ from each corner and folding, the resulting dimensions of the box will be:
     • Length: $9 - 2x$
     • Width: $8 - 2x$
     • Height: $x$

2. Account for the double layer along one side:

   • Since one of the longer sides will have a double layer, this does not change the base area of the box or the height $x$. It affects only the construction, not the volume.

3. Write the volume formula:

   • The volume $V$ of the box is given by: $V = \text{length} \times \text{width} \times \text{height}$ Substituting the expressions we have: $V = (9 - 2x)(8 - 2x)x$ Expanding this: $V = (72 - 34x + 4x^2)x = 4x^3 - 34x^2 + 72x$

4. Maximize the volume by finding the critical points:

   • Take the derivative of $V$ with respect to $x$ and set it equal to zero to find the critical points: $\frac{dV}{dx} = 12x^2 - 68x + 72$ Setting this equal to zero: $12x^2 - 68x + 72 = 0$ Simplify by dividing by 4: $3x^2 - 17x + 18 = 0$

5. Solve the quadratic equation:

   • Using the quadratic formula: $x = \frac{-(-17) \pm \sqrt{(-17)^2 - 4 \cdot 3 \cdot 18}}{2 \cdot 3}$ $x = \frac{17 \pm \sqrt{289 - 216}}{6}$ $x = \frac{17 \pm \sqrt{73}}{6}$
   • Approximating, we get two values for $x$: $x \approx 3.434$ and $x \approx 1.733$.

6. Evaluate the feasible solution:

   • Since $x = 3.434$ would make the box dimensions negative, we discard it.
   • Therefore, $x \approx 1.733$ is feasible.

7. Calculate the maximum volume:

   • Substitute $x = 1.733$ into the volume formula: $V = 4(1.733)^3 - 34(1.733)^2 + 72(1.733)$
   • Evaluating this gives: $V \approx 43.65 \text{ cubic feet}$

Answer:

The largest possible volume for the box is approximately 43.65 cubic feet.

Would you like more details on any step?

Here are 5 questions to consider:

1. How would the volume change if the cardboard dimensions differed?
2. What if we added a constraint to limit the height to a maximum value?
3. How would we adjust if two sides required a double layer?
4. How does the choice of $x$ affect the strength of the box?
5. What happens to the volume if we eliminate the double-layer constraint?

Tip: Always double-check constraints to confirm they don’t change the optimization process for volume or area.